Same question idea but different values... I hope I helped you... Don't forget to put a heart mark
if a voltage is applied to a capacitor, current flows easily at first and then slows as the capacitor becomes charged. Inductors
behave just the opposite, in that they reluctantly pass current when a voltage is first applied, and then the current passes easily as time passes. If the input voltage is suddenly raised from zero to some constant value, sketch the current in the capacitor, iC, and the inductor, iL, as a function of time. What is the steady-state current in the capacitor and inductor?
1 answer:
Answer: photo!
Explanation:
For the capacitor, the current as a function of time will be
I(t)=V/R e^(-t/RC)
In steady-state, time tends to infinite, so current tends to zero.
For the inductor, the current as a function of time is
I(t)=V/R (1-e^(-tR/L))
In steady-state, time tends to infinite, so current tends to V/R
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Answer:
the critical flaw length is 10.06 mm
Explanation:
Given the data in the question;
plane strain fracture toughness = 92 Mpa√m
yield strength σ = 900 Mpa
design stress is one-half of the yield strength ( 900 Mpa / 2 ) 450 Mpa
Y = 1.15
we know that;
Critical crack length = 1/π(
/ Yσ )²
we substitute
= 1/π( 92 Mpa√m / (1.15 × 450 Mpa )²
= 1/π( 92 Mpa√m / (517.5 Mpa )²
= 1/π( 0.177777 )²
= 1/π( 0.03160466 )
= 0.01006 m = 10.06 mm
Therefore, the critical flaw length is 10.06 mm
{ = ( 10.06 mm ) > 3 mm
The critical flow is subject to detection